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在矩形abcd中,点e在ad上,ec平分角BED若ab=1角abe=45度求AD长
人气:267 ℃ 时间:2020-04-13 04:21:11
解答
∠BED = ∠BAE+∠ABE = 90°+45° = 135° ,
∠DCE = 90°-∠CED = 90°-(1/2)∠BED = 22.5° ,
tan22.5° = tan(45°/2) = (1-cos45°)/sin45° = √2-1 ,
CD = AB = 1 ,
AE = AB*tan∠ABE = AB*tan45° = 1 ,
DE = CD*tan∠DCE = CD*tan22.5° = √2-1 ,
AD = AE+DE = √2 .
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